From 4cyclenormal, we see that (perhaps after permuting the generators), we have either \(2a = 2b\)  or \(b = 3a\)  or \(4a = 0\) . However, we cannot have \(2a = 2b\) , since \(X\)  is twin-free. So there are two cases to consider.
Case 1. Assume \(b = 3a\) .
Since \(|b| \ne 2\) , we have \(n > 6\) .
For \(n = 8\) , we have \(2b = 6a = -2a\) , which contradicts the assumption that \(X\)  is twin-free.
For \(n = 10\) , we have the Cayley graph that is specified in the statement of the BipNonnormalVal4.
For \(n \ge 12\) , we have \(X \cong \operatorname{Cay}(\mathbb {Z}_n, \lbrace \pm 1, \pm 3\rbrace )\) . This is the situation of [1]}, but, for completeness, we sketch the proof.
The only nontrivial automorphism of the ball of radius 2 centered at 0 is \(x \mapsto -x\) . From this, it easily follows that \(x \mapsto -x\)  is the only nontrivial automorphism of \(X\) , so \(X\)  is normal. This contradicts our hypothesis.
Case 2. Assume \(4a = 0\) .
This means \(|a| = 4\) . Since \(X\)  is bipartite, we know that \(|b|\)  is even, so \(\langle b \rangle \)  contains the unique element of order 2; this means \(2a = \ell b\)  for some \(\ell \in \mathbb {Z}\) . Since \(X\)  is bipartite, we know that \(\ell \)  is even. So \(|b|\)  is divisible by 4. Therefore, \(\langle b \rangle \)  contains \(a\) . So \(X \cong \operatorname{Cay}(n, \lbrace \pm 1, \pm n/4\rbrace )\) . If we consider the subgraph induced by the ball of radius 2, we note that only the vertices \(\pm 1\)  have a pendant edge. In particular, every automorphism of \(X\)  maps 1-edges to 1-edges and is therefore an automorphism of the graph \(X_0\operatorname{Cay}(\mathbb {Z}_n,\lbrace \pm 1\rbrace ) \cong C_n\) . Since this Cayley graph is normal, we can conclude the same about \(X\) , which is a contradiction.
